Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{6(2r + 5)}{4r} \times \dfrac{8}{18r + 45} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $n = \dfrac{ 6(2r + 5) \times 8 } { 4r \times (18r + 45) } $ $ n = \dfrac {8 \times 6(2r + 5)} {4r \times 9(2r + 5)} $ $ n = \dfrac{48(2r + 5)}{36r(2r + 5)} $ We can cancel the $2r + 5$ so long as $2r + 5 \neq 0$ Therefore $r \neq -\dfrac{5}{2}$ $n = \dfrac{48 \cancel{(2r + 5})}{36r \cancel{(2r + 5)}} = \dfrac{48}{36r} = \dfrac{4}{3r} $